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प्रश्न
Factorise the following, using the identity a2 – 2ab + b2 = (a – b)2.
p2y2 – 2py + 1
योग
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उत्तर
We have,
p2y2 – 2py + 1
= (py)2 – 2 × py × 1 + 12
= (py – 1)2
= (py – 1)(py – 1)
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अध्याय 7: Algebraic Expression, Identities and Factorisation - Exercise [पृष्ठ २३४]
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